ISI2016-DCG-63

Question

If $\sin^{-1}\frac{1}{\sqrt{5}}$ and $\cos^{-1}\frac{3}{\sqrt{10}}$ lie in $\left[0,\frac{\pi}{2}\right],$ their sum is equal to

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{3}$
• C. $\sin^{-1}\frac{1}{\sqrt{50}}$
• D. $\frac{\pi}{4}$

Comments

@ankitgupta.1729

Can you please tell how this question can be solved?

---

@`JEET

Assume, $\alpha = sin^{-1}\frac{1}{\sqrt{5}}$ and $\beta = cos^{-1}\frac{3}{\sqrt{10}}$

Now, we have to find the values of $\alpha + \beta$.

Since, $\alpha = sin^{-1}\frac{1}{\sqrt{5}}$. So, $sin{\alpha} = \frac{1}{\sqrt{5}}$ and $cos{\alpha} = \frac{2}{\sqrt{5}}$

Similarly, $\beta = cos^{-1}\frac{3}{\sqrt{10}}$. So,  $cos{\beta} = \frac{3}{\sqrt{10}}$ and $sin{\beta} = \frac{1}{\sqrt{10}}$

Now, use the formula of $sin(\alpha + \beta) $ and find $\alpha + \beta$. Answer should be (D)

---

Thanks.

Between, I was confused whether its $\sin \alpha + \cos\alpha ~ \text {or} ~ \alpha + \beta $.

Answer 1

Answer: $\textbf D$

Let: $\bf {\alpha} = \sin ^{-1}\frac{1}{\sqrt 5}$, and $\bf {\beta} = \cos^{-1} \frac{3}{\sqrt {10}}$


$\therefore\mathbf{ \sin \alpha} = \frac{1}{\sqrt 5} = \frac{\bf P}{\bf H}, \; \cos \beta = \frac{3}{\sqrt{10}}= \frac{\bf B} {\bf H}$

By using Pythagoras Theorem:

$(\because \mathrm {P^2 + B^2 = H^2 \Rightarrow B = \sqrt {H^2 - P^2}} = \sqrt {5-1} = 2)$

$\Rightarrow \cos \alpha = \frac{2}{\sqrt 5}$

Similarly,

$\Rightarrow \cos \beta = \frac{3}{\sqrt {10}}$

$\Rightarrow \sin \beta = \frac{1}{\sqrt {10}}$

Now, We know that:

$\mathrm {\sin (A+B) = \sin A \cos B + \cos A \sin B}$

$\begin {align}\Rightarrow \sin(\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta\\ &= \frac{1}{\sqrt 5}.\frac{3}{\sqrt {10}} + \frac{2}{\sqrt {5}}. \frac{1}{\sqrt{10}} \\ &= \frac{3}{\sqrt{50}} + \frac{2}{\sqrt{50}}\\&=\frac{5}{\sqrt{50}}\\&= \frac{5}{5\sqrt 2} \\&=\frac{1}{\sqrt 2}\end {align}$

$\begin {align} \Rightarrow \alpha + \beta &= \sin ^{-1} \frac{1}{\sqrt 2} \\&= \sin ^{-1} \Bigg({\sin \frac{\pi}{4}}\Bigg) \\&= \frac{\pi}{4}\end {align}$

Answer $\therefore \textbf D $ is the correct option.

Comments

Perfect one.

---

Thanks!